Question

A 20.00-ml sample of 0.3000 m hbr is titrated with 0.15 m naoh. what is the ph of the solution after 40.3 ml of naoh have been added to the acid?

Answer 1

$Molarity = no.of moles/vol.of the solution$

Then, $no.of moles of NaOH= 0.15M*40.3ml$

$= 6.04510x^{-3} moles$

$no.of moles of HCl= 0.30M×20ml= 6.010^-3 moles$

The reaction is as bellows:

$HCl+NaOH= NaCl+H2O$

1 mole$HCl$ reacts with 1 mole $NaOH$

Therefore,$6.0 × 10^- .. 6.0 × 10 ^-3 moles of NaOH$

Then, unreacted $NaOH= (6.045×10^-3–6.0×10^-3) moles$

$= 4.5×10^-5 moles$

Volume of the solution = $(20+40.3)ml= 60.3ml= 6.03×10–2 L$

Now, molar concentration of $NaOH= 4.5×10^-5 moles/6.03×10^-2L$

$= 7.46×10–4 moles/L$

$NaOH$ completely dissociates in solution.

Therefore,$[NaOH]=[[OH-]=7.46×10^-4 moles/L$

$pOH= -log[OH-]= -log(7.46×10^-4)= 3.12$

Again,$pH+pOH=14$

or,$pH=14-pOH$

$= 14–3.12= 10.88$

If sodium hydroxide were added to a solution of strong acid, what would happen to the pH of the solution?

Depending on how much $NaOH$ is added, yes. The solution has a pH of 7 if the amount of additional $NaOH$ is exactly equal to the amount of acid.

Since $NaOH$ is an excess reagent, the solution will become basic if more equivalents of base are added than equivalents of acid, and the pH of the solution will rise above 7.

The solution will be acidic, with a pH lower than 7, if the number of equivalents of acid exceeds that of the $NaOH$ that has been added.

To learn more about Sodium hydroxide,visit:

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